Question

(a) What is the oxidation state of \(\mathrm{P}\) in \(\mathrm{PF}_{6}^{-}\) and of \(\mathrm{N}\) in \(\mathrm{NF}_{3} ?\) (b) Why doesn't \(\mathrm{N}\) form \(\mathrm{NF}_{6}^{-}\) ion analogous to \(\mathrm{P} ?\)

Short answer

The oxidation state of P in PF6- is +5, and the oxidation state of N in NF3 is +3. Nitrogen does not form an NF6- ion because it belongs to the 2nd period and has only three unpaired electrons in its 2p orbital. Unlike phosphorus, nitrogen does not have access to empty d orbitals to accommodate extra electrons and extend its octet, allowing it to form only three bonds with three fluorine atoms in NF3.

Step-by-step solution

Determine the oxidation state of P in PF6-

To determine the oxidation state of P in PF6-, we have to note that the oxidation state of F is always -1. Since in the compound PF6-, there are 6 fluorine atoms, the total negative charge is -6. Since PF6- ion has a total negative charge of -1, the oxidation state of P has to be +5 in order to achieve the net charge of -1. Therefore, the oxidation state of P in PF6- is +5.

Determine the oxidation state of N in NF3

To determine the oxidation state of N in NF3, we have to consider that the oxidation state of F is -1. Since there are 3 fluorine atoms, the total negative charge is -3. Nitrogen, being the more electronegative element, has a lower oxidation state, which must be +3 in order to balance the net charge of the compound. Therefore, the oxidation state of N in NF3 is +3.

Analyze the electronic configuration of P and N

In order to understand why N doesn't form an NF6- ion analogous to P, let's analyze their electronic configurations. The electronic configuration of N is 1s² 2s² 2p³, and that of P is 1s² 2s² 2p⁶ 3s² 3p³. Nitrogen belongs to the 2nd period and has three unpaired electrons in the 2p orbital. On the other hand, phosphorus belongs to the 3rd period and has three unpaired electrons in the 3p orbital. However, phosphorus also has empty 3d orbitals, which can be used for the expansion of its octet by forming additional bonds. This is not the case for nitrogen, as it lacks available empty d orbitals for extra bond formation.

Explain why N doesn't form NF6- ion

Based on the electronic configuration analysis, nitrogen doesn't form the NF6- ion because it can only make three bonds with its three unpaired electrons in the 2p orbital. Since nitrogen belongs to the 2nd period, it doesn't have access to empty d orbitals to accommodate extra electrons that could extend its octet, as phosphorus does. Nitrogen can form three bonds with three fluorine atoms, resulting in an NF3 molecule, but it cannot form additional bonds, like phosphorus, to create an NF6- ion. In conclusion, the oxidation state of P in PF6- is +5, the oxidation state of N in NF3 is +3, and N doesn't form NF6- ion due to its electronic configuration preventing the extension of its octet.

Key concepts

Electronic Configuration

Electronic configuration is crucial in understanding how elements form compounds. It refers to the arrangement of electrons in the orbitals of an atom. Each element has a unique configuration that determines its chemical behavior. For instance, nitrogen has the electronic configuration of 1s² 2s² 2p³. This setup reveals that nitrogen has five valence electrons, with three unpaired in the 2p orbital, ready for bond formation.
Phosphorus, on the other hand, has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p³. It belongs to the third period, and its configuration shows availability of five valence electrons. Unlike nitrogen, phosphorus has vacant 3d orbitals. These can potentially allow phosphorus to expand its usual octet when bonding, a crucial difference that limits nitrogen's bonding capability.

Phosphorus Compounds

Phosphorus is versatile in forming compounds due to its ability to expand its octet. This characteristic is primarily due to the availability of its empty 3d orbitals. One classic example is in compounds like \( \text{PF}_6^− \).
In \( \text{PF}_6^− \), phosphorus achieves an oxidation state of +5. This is because each fluorine atom, with an oxidation state of -1, pulls electrons away from phosphorus. With six fluorine atoms in the compound, phosphorus must have an oxidation state of +5 to balance the −6 charges from fluorine, leading to the overall charge of −1 on the molecule.

Nitrogen Compounds

Nitrogen forms stable compounds too, but with some constraints due to its electronic configuration. In the case of \( \text{NF}_3 \), nitrogen exhibits an oxidation state of +3.
This is because, even though it has the capacity to form three bonds with three fluorine atoms, it doesn't have the capability to expand its octet. Nitrogen lacks empty d orbitals in its valence shell, restricting its ability to accommodate more than eight electrons, unlike phosphorus.

Octet Rule

The octet rule is fundamental in understanding the stability of chemical compounds. It states that atoms tend to gain, lose, or share electrons to achieve a full set of eight valence electrons, resembling the electron configuration of noble gases.
Both nitrogen and phosphorus aim to follow this rule, but phosphorus has a unique capability to expand its octet due to available 3d orbitals. Nitrogen, limited to the second energy level, strictly adheres to the octet rule, forming compounds with a maximum of three bonds.

Bond Formation

Bond formation is essential for creating stable compounds. It involves the sharing or transfer of electrons between atoms. The nature of the electronic configuration of an element largely determines how bonds are formed.
Nitrogen can form three covalent bonds in compounds such as \( \text{NF}_3 \) due to its three unpaired electrons in the 2p orbital. Phosphorus, in contrast, can form more than three bonds, as seen in \( \text{PF}_6^− \), taking advantage of its ability to expand the octet due to the presence of empty d orbitals. This ability to form additional bonds allows phosphorus to create a wider variety of compounds than nitrogen.